Hilbert Symbol over $\mathbb{R}$ (bilinearity)

Let $\mathbb{R}$ be the field of the reals and let $a,b,c \in \mathbb{R}^{\times}$. As you probably know, the Hilbert symbol over any field $K$ is defined as: $$(\frac{a,b}{K}) = 1 \text{ if } \exists x,y \in K \text{ such that } ax^2 + by^2 = 1,$$ and $-1$ otherwise. I've proven the multiplicative result $$(\frac{a,bc}{\mathbb{Q}_p}) = (\frac{a,b}{\mathbb{Q}_p})(\frac{a,c}{\mathbb{Q}_p})$$ for the field $\mathbb{Q}_p$, but can't find the reasoning for $\mathbb{R}$, strangely enough. I have the feeling that it's easily verified, but I'm not seeing it. Any thoughts?

Thanks!

If you work with the real numbers then the existence of $x,y$ satisfying $ax^2+by^2=1$ only depends on wether one of $a,b$ is nonnegative. For example if $a>0$ then $(x,y)=\left(\frac{1}{\sqrt{a}},0\right)$ is a possible solution.
With that in mind it becomes a simple discussion on signs : if $a>0$ then all terms in your equation are $1$, and if not then $\left(\frac{a,b}{\mathbb{R}}\right)$ is the sign of $b$, so the multiplicative property holds.